Question

After rejecting the null hypothesis of equal treatments, a researcher decided to compute a 95 percent confidence interval for the difference between the mean of treatment 1 and mean of treatment 2 based on the Tukey procedure. At α = .05, if the confidence interval includes the value of zero, then we can reject the hypothesis that the two population means are equal.

True

False

The error sum of squares measures the between-treatment variability.

True

False

The experimentwise α for the 95 percent individual confidence interval for μ₁ − μ₂ (treatment mean 1 − treatment mean 2) will always be smaller than the experimentwise α for a Tukey 95 percent simultaneous confidence interval for μ₁ − μ₂.

True

False

In one-way ANOVA, a large value of F results when the within-treatment variability is large compared to the between-treatment variability.

True

False

Answer 1

False , this is not true that After rejecting the null hypothesis of equal treatments, a researcher decided to compute a 95 percent confidence interval for the difference between the mean of treatment 1 and mean of treatment 2 based on the Tukey procedure. At α = .05, if the confidence interval includes the value of zero, then we can reject the hypothesis that the two population means are equal.

True, The error sum of squares measures the between-treatment variability.

this is True , The experimentwise α for the 95 percent individual confidence interval for μ₁ − μ₂ (treatment mean 1 − treatment mean 2) will always be smaller than the experimentwise α for a Tukey 95 percent simultaneous confidence interval for μ₁ − μ₂.

true ,In one-way ANOVA, a large value of F results when the within-treatment variability is large compared to the between-treatment variability.